You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in the example below.

Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.

Example 1:

Input: head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
Output: [1,2,3,7,8,11,12,9,10,4,5,6]
Explanation:

The multilevel linked list in the input is as follows:
Flatten a Multilevel Doubly Linked List
After flattening the multilevel linked list it becomes:
Flatten a Multilevel Doubly Linked List

Example 2:

Input: head = [1,2,null,3]
Output: [1,3,2]
Explanation:
The input multilevel linked list is as follows:
1---2---NULL
|
3---NULL

Example 3:

Input: head = []
Output: []

How multilevel linked list is represented in test case:

We use the multilevel linked list from Example 1 above:

 1---2---3---4---5---6--NULL
|
7---8---9---10--NULL
|
11--12--NULL

The serialization of each level is as follows:

[1,2,3,4,5,6,null]
[7,8,9,10,null]
[11,12,null]

To serialize all levels together we will add nulls in each level to signify no node connects to the upper node of the previous level. The serialization becomes:

[1,2,3,4,5,6,null]

Constraints:

  • The number of Nodes will not exceed 1000 .
  • 1 <= Node.val <= 105

Solution:

class Solution:
def flatten_dfs(self, head):
curr = head
prev = None
while curr:
if curr.child:
succ = curr.next 
# Get head and tail of linked list on next level 
child_head, child_tail = self.flatten_dfs(curr.child)
curr.next = child_head
curr.child = None
child_head.prev = curr
child_tail.next = succ
curr = succ
# Case when the last node on the level has a child (succ node is None)
if succ:
succ.prev = child_tail
prev = succ.prev
else:
prev = child_tail
else:
prev = curr
curr = curr.next
return head, prev
def flatten(self, head: 'Node') -> 'Node':
return self.flatten_dfs(head)[0] 
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Categories: InterviewPython

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